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For any natural number $a ∈ N$ , the exponential map of base $a$ is the map $a^ {( )} \mathbb{N} \rightarrow \mathbb{N}$, $n \mid \rightarrow a^n$,

defined recursively (using the recursion theorem) by setting

$a^0 := 1$, and for any $n ∈ \mathbb{N}$, $a^{n+1} := (a^n)(a)$

(a) Show that for any $n,m ∈ N$, one has $(a^n)(a^m) = a^{n+m}$

(b) Show that for any $n,m ∈ \mathbb{N}$, one has $(a^n)^m = a^{nm}$

Seeking assistance on this question , totally clueless. Much thanks!

Joe
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  • Use the definition of the exponentiation map to write each expression as a product of many $a$s, and count how many $a$s there are. – angryavian Sep 25 '14 at 16:28

1 Answers1

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For Problem (a), I suggest fixing $n$ and using induction on $m$. The case $m=0$ is easy.

For the induction step, we wish to show that if $(a^n)(a^k)=a^{n+k}$, then $(a^n)(a^{k+1})=a^{n+(k+1)}$. We have $$(a^n)(a^{k+1})=(a^n)((a^k)(a))=((a^n)(a^k))(a).\tag{1}$$ But by the induction hypothesis we have $(a^n)(a^k)=a^{n+k}$, and by definition $(a^{n+k})(a)=a^{(n+k)+1}=a^{n+(k+1)}$.

We leave the induction proof of (b) to you. Use again the strategy of fixing $n$. In the induction step, (a) will be useful.

André Nicolas
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  • You are welcome. I hope that the proof for $(a^n)^m$ went smoothly. – André Nicolas Sep 27 '14 at 17:03
  • i fix n , induct on m. Basically, i let p(m) be the statement for (a^n)^m = a^(n.m) for a , n , m ∈ R. I let m = 0, i prove P(0). Then i assume p(k) is true. Then i show p(k+1) is true. For p(k+1), heres what i did. (a^n)^(k+1) = a^(nk).a^(n) [by definition) = (a^n)^k . a^n [by induction hypothesis] = a^(n(k+1)) [by induction hypothesis on (a) indeed] and yes! i conclude that p(k) is true , p(k+1) is true, p(0) is true. Therefore i show that indeed (b) results is true. Sorry for the messy equations! – Joe Sep 27 '14 at 17:31
  • We need some change, the order above is not right. We have $(a^n)^{k+1}=(a^n)^k a^n$ by definition. Then $(a^n)^k=a^{nk}$ by induction hypothesis. So we reach $a^{nk}a^n$. By the first part of the problem (done in detail in the answer) this is $a^{nk+n}$. This is $a^{n(k+1)}$. – André Nicolas Sep 27 '14 at 17:40
  • Hi, by the first part of the problem , i just mention : (by the results from part (a) ) will suffice right? Thanks for pointing out the order! Careless on my part! Now i just need to conclude with the mathematical induction statement at the end (p(k) is true, p(k+1) is true ...... then (b) results is true. Thank you andre for the help! – Joe Sep 27 '14 at 17:48
  • You are welcome. When we are giving formal proofs of results we have used since childhood, it can be hard to separate what we have proved from what we "know" to be true. – André Nicolas Sep 27 '14 at 17:53
  • Dear andre, is it possible if you help me on the question i posted regarding injective when ϕ is post-composed to f? – Joe Sep 28 '14 at 13:13